Codable class does not conform to protocol Decodable

Question

Why am I getting a "Type 'Bookmark' does not conform to protocol 'Decodable'" error message?

class Bookmark: Codable {
   weak var publication: Publication?
   var indexPath: [Int]
   var locationInText = 0

   enum CodingKeys: String, CodingKey {
      case indexPath
      case locationInText
   }

   init(publication: Publication?, indexPath: [Int]) {
      self.publication = publication
      self.indexPath = indexPath
   }
}

I do not wish to save the publication var since the Publication owns the Bookmark but the bookmark needs to know which Publication it belongs to. The decode init of Publication will set the bookmark reference to itself.

Cannot reproduce in Swift 4.0.3 (Xcode 9.2) – please could you provide a [mcve]? — Hamish, Feb 01 '18 at 17:29
It would be nice to know why my question was downvoted so that I could actually learn something for the future — Melodius, Feb 01 '18 at 18:38
Not one of the downvoters, but your question would be much more useful with a self-contained example so that others can easily reproduce your problem (I still can't reproduce it). — Hamish, Feb 01 '18 at 18:41

score 69 · Answer 1 · answered Feb 01 '18 at 17:27

69

The compiler cannot synthesise the required init(from:) method due to the weak reference, so you need to write it yourself.

class Bookmark: Codable {
    weak var publication: Publication?
    var indexPath: [Int]
    var locationInText = 0

    private enum CodingKeys: String, CodingKey {
        case indexPath
        case locationInText
    }

    init(publication: Publication?, indexPath: [Int]) {
        self.publication = publication
        self.indexPath = indexPath
    }

    required init(from decoder:Decoder) throws {
        let values = try decoder.container(keyedBy: CodingKeys.self)
        indexPath = try values.decode([Int].self, forKey: .indexPath)
        locationInText = try values.decode(Int.self, forKey: .locationInText)
    }
}

answered Feb 01 '18 at 17:27

Dávid Pásztor

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Your answer was correct too but matt was first. I don't know what the proper thing to do here is, but enlighten me if needed. Thanx. – Melodius Feb 01 '18 at 18:31
Glad I could help. It's up to you, which answer you accept, but if you found both useful, you can still upvote both regardless of which one do you accept. – Dávid Pásztor Feb 01 '18 at 18:34
@Melodius up to you to choose which answer you think is best and would provide the most information to other people with the same question – ColdLogic Mar 29 '18 at 19:06
8

I found David's answer more helpful since it has sample code. – Lone Ronin Jun 25 '18 at 18:42
@DávidPásztor I'm wondering why `publication` has anything to do with encoding/decoding of `Bookmark`, since it's excluded from the `CodingKeys`? – jeff-h Jan 16 '20 at 09:26
auto-generated decoding requires all variables to be initialized afterward, instead of implementing `init(from:)` you can set a default value for variables not in the `CodingKeys` - `weak var publication: Publication? = nil` – bshirley Nov 28 '21 at 16:38

matt · Accepted Answer · 2018-02-01T17:36:27.060

50

Why am I getting a "Type 'Bookmark' does not conform to protocol 'Decodable'" error message

It's either because Publication isn't Decodable (you have not shown what it is, so it's hard to tell) or because of the weak designation on publication.

Either way, it's easy to fix: you just need to implement init(from:) to complete the implementation of Decodable; the compiler is simply telling you that this implementation cannot be synthesized.

edited Feb 01 '18 at 17:36

answered Feb 01 '18 at 17:22

matt

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I believe you will then have a parallel issue for the Encodable side, but let's fix one thing at a time. – matt Feb 01 '18 at 17:28
Works now thanks! Encodable side did not complain. Just curious as to why the compiler is perfectly happy if I remove CodingKeys altogether when I haven't implemented init(from:)? – Melodius Feb 01 '18 at 18:29
1

Publication didn't have anything to do with the problem this time. – Melodius Feb 01 '18 at 18:29
It's not decodable because `publication` isn't initialized after the object is decoded. `weak var publication: Publication? = nil` Provide a default value and it be become `Decodable`. – bshirley Nov 28 '21 at 16:36
@bshirley If you think you have a better answer, please do give it as an actual answer. – matt Nov 28 '21 at 17:20
My problem was working with SwiftUI and having one variable inside my class marked as @Publishable. Everything conforms to Codable, but the compiler still choked. Conformed manually, everything works fine. – green_knight May 01 '22 at 23:33
Note: This can also happen when CodingKeys does not define keys for all variables in the struct. Especially, when some keys do not fit variable names at all. – Fabian Sep 10 '22 at 13:27

score 36 · Answer 3 · answered Sep 04 '18 at 19:34

36

Another reason you could get this message is if your CodingKeys enum isn't exhaustive. If you have three properties in the data type, then your CodingKeys enum needs to have three property/name cases as well.

answered Sep 04 '18 at 19:34

mikepj

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Also check typos in CodingKeys. If there are only one different character, code wont compile. (It was my case, thanks @mikepj) – Pavel Shorokhov Nov 13 '18 at 15:40
3

But surely the whole point of specifying your own `CodingKeys` is because you want to exclude one or more properties from being encoded? – jeff-h Jan 16 '20 at 09:22
I was off by a single char, I was about to rip my hair out ty! – Hajji Daoud Jan 15 '21 at 22:11
3

You **do NOT need to exhaustively list your properties in the CodingKeys**. If you do not list a property, you need to provide a default value for it. – bshirley Nov 28 '21 at 16:28
That was it in my case. Thanks. – ervinbosenbacher Apr 25 '23 at 19:03

iOSer · Answer 4 · 2019-03-29T05:27:56.920

33

On the hindsight, I received a similar error when trying to set Codable to my class which consisted of NSNumber type variables. See image below:

Changing NSNumber to primitive data type Int resolved the issue. See below:

I'm guessing this might be true for other datatypes that require bridging to Swift Standard Library Value Types such as NSString, NSArray and so on

edited Mar 29 '19 at 05:27

answered Apr 05 '18 at 09:35

iOSer

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Thank you! In my case I was trying to use NSNumber. Switched to Int per your suggestion and bam it works – Sleeping_Giant Nov 14 '18 at 05:56
1

@Sleeping_Giant Haha. You're Welcome. As always happy to help :) – iOSer Nov 14 '18 at 06:10

score 31 · Answer 5 · answered Apr 13 '21 at 04:12

31

In a similar scenario, I was getting the same issue because the variable name in my CodingKeys was not the same as a class variable. See below

answered Apr 13 '21 at 04:12

Amit Baderia

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easy to make this mistake, thanks. I almost went down the custom init route, then saw your answer – Morten J Feb 07 '22 at 15:16
wow, just made this mistake. Thanks – Simon McNeil Feb 18 '23 at 15:06

Wissa · Answer 6 · 2019-05-30T10:06:05.820

23

Simply because your CodingKeys enum is not exhaustive, add publication property to the enum to achieve that.

try this:

class Bookmark: Codable {
   weak var publication: Publication?
   var indexPath: [Int]
   var locationInText = 0

   // All your properties should be included
   enum CodingKeys: String, CodingKey {
      case indexPath
      case locationInText
      case publication   // this one was missing
   }
}

You wont need the init method anymore as the implementation now can be synthesized.

edited May 30 '19 at 10:06

answered May 30 '19 at 09:59

Wissa

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weak may only be applied to class and class-bound protocol types – Minkoo Dec 09 '19 at 11:11
Thank you I had a codable struct under a codable struct and I forgot to add the inner struct as the coding key. – Natasha Nov 25 '22 at 09:56

bikram sapkota · Answer 7 · 2021-09-30T03:57:51.823

13

Any class to be a codeable must have it's all property codeable. Standard library types like String, Int, Double and Foundation types like Date, Data, and UR confirm the codeable protocol but some doesn't.

For e.g below Note class have all properties of string which confirm codable protocol so no error:

But UIImage don't confirm codable protocol so it throw error:

edited Sep 30 '21 at 03:57

answered Aug 21 '20 at 04:29

bikram sapkota

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This was very helpful. I had a custom class as a property that was not codable. – DevinM Sep 15 '21 at 12:36

score 4 · Answer 8 · answered Jun 03 '19 at 10:43

You can omit a property from coding keys enum, only if it has a default value.

From apple docs

Omit properties from the CodingKeys enumeration if they won't be present when decoding instances, or if certain properties shouldn't be included in an encoded representation. A property omitted from CodingKeys needs a default value in order for its containing type to receive automatic conformance to Decodable or Codable.

score 3 · Answer 9 · edited May 16 '19 at 11:55

3

In-short, while implementing Codable, all properties which are non-primitive data type (mean class type or may it can be objective-c class) must be Codable.

   weak var publication: Publication?

in this case publication is of type class so Publication must have implemented Codable

edited May 16 '19 at 11:55

pkc456

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answered Jan 03 '19 at 08:09

kalpesh jetani

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weak' may only be applied to class and class-bound protocol types – Minkoo Dec 09 '19 at 11:11

score 2 · Answer 10 · answered May 26 '20 at 10:23

2

Bit of a daft one but in case it helps someone else. I got this error because I put enum CodingKeys: CodingKeys instead of enum CodingKeys: CodingKey.

answered May 26 '20 at 10:23

Declan McKenna

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Very helpful, I had done the same mistake – ummer akbar Jan 23 '23 at 11:27

Amisha Italiya · Answer 11 · 2023-02-14T09:07:43.610

If you have tried all the above solutions and are still unable to fix the error then I think it could be because of the data type you are using for your data class fields.

In the question they have used 1 field weak var publication: Publication?, and if you have the same kind of class structure then maybe you should check if that Publication data class conforms to the Codable class or not.

Because it is mandatory to conform same protocol for the child class as well, as it's fields are also should be encodable and decodable.

score 0 · Answer 12 · answered Aug 16 '18 at 21:15

0

I had a similar issue which I stumbled upon this fix to. As I am new to Swift, I am unsure as to why it works! If anyone knows I would appreciate the knowledge.

I changed this:

let id, type: Int

to this:

let id: Int
let type: Int

answered Aug 16 '18 at 21:15

ThunderGrad

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Codable class does not conform to protocol Decodable

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